How to show $|\partial_x^n\partial_u^k p(u,x)| \leq Cu^{-|n|/2-k}p(cu,x)$; $p(u,x)=\frac{1}{(4\pi u)^{q/2}}e^{-|x|^2/(4u)}$ is the normal distribution Let $p(u,x)=\frac{1}{(4\pi u)^{q/2}}e^{-|x|^2/(4u)},u>0,x \in \mathbb{R}^q.$

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*Prove that for $r\geq 0,c>1$ there exists $C>0$ (depending on $r,c$) such that $$\forall x \in \mathbb{R}^q,u>0,\frac{|x|^{2r}}{u^r}p(u,x) \leq Cp(cu,x).$$

*Deduce that for $n \in \mathbb{N}^q,k \in \mathbb{N},$ there exists $C'>0$ such that $|\partial_x^n\partial_u^k p(u,x)| \leq C' u^{-|n|/2-k}p(cu,x)$ where $\partial^n_x=\partial x_1^{n_1}...\partial x_q^{n_q}$
Attempt:

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*We have $\frac{|x|^{2}}{4ur}(1-\frac{1}{c}) \leq e^{\frac{|x|^2}{4ur}(1-1/c)}$ and we choose $C=c^{q/2} 4^r(r+1)^r/(1-1/c)^r$
How to deduce 2 (for simplicity we can take $q=1)$? Is it possible to prove the inequality using induction? If so, how?
 A: For 1:$$\frac{|x|^{2r}}{u^r} p(u,x)=\frac{|x|^{2r}}{u^r}\frac{1}{(4\pi u)^{q/2}}e^{-|x|^2/(4u)}, \qquad  p(cu,x)=\frac{1}{(4\pi u)^{q/2}c^{q/2}}e^{-|x|^2/(4cu)}, $$
so we are being asked to show that there exists $C=C(r,c,q)$ where
$$ \frac{|x|^{2r}}{u^r} \le \frac C{c^{q/2}}e^{\frac{-|x|^2}{4u}(\frac1c-1)}$$
Note $1/c-1<0$ so for $r=0$ we can just take $C=c^{q/2}.$
For $r$ a positive integer we have for $A>0,X\ge0$ that  $e^{AX} \ge A^rX^r/r!$. Applying this with $A = -(1/c-1)/4$ and $X=|x|^2/u$ we get
$e^{\frac{-|x|^2}{4u}(\frac1c-1)}\ge \big(\frac{1-1/c}4\big)^r \frac{|x|^{2r}}{r!u^r}$ so we can take $$C = \frac{c^{q/2}r!}{\big(\frac{1-1/c}4\big)^r}$$
This implies estimates for all $r\ge0$ by interpolating which I will leave to you.
Idea for 2: use induction to show that $\partial_x^n\partial_u^k p(u,x) = \sum_{i,j} x^{a_i} u^{b_j} p(u,x)$ for some appropriate powers $a_i,b_j$
then apply 1.
A: It is quite hard to prove the inequality directly with induction. This is because you cannot differentiate an inequality; $f\le g$ does not imply that $f' \le g'$.
Morally, you need to induct on something which you can differentiate.
In my earlier answer I provided the following hint:

use induction to show that $\partial_x^n\partial_u^k p(u,x) = \sum_{i,j} x^{a_i} u^{b_j} p(u,x)$ for some appropriate powers $a_i,b_j$
then apply 1.

I got the reply "For 2 it seems there is a lot of work to do considering it to be deduced from 1". This Answer hopefully shows that this isn’t really the case, especially in the age of chatGPT WolframAlpha. Checking small cases like this, not to mention being given the wanted result, we can refine our conjecture to

For each $\alpha\in \mathbb N^q,k\in\mathbb N$, the expression $\partial_x^\alpha \partial_u^k p(u,x)$ can be written as a sum of finitely many (depending on $q,k,\alpha$) terms of the form $x^{\beta}u^{-|\beta|/2-|\alpha|/2-k} p(u,x)$

Which clearly implies the required result by absorbing the $\beta$ terms using part 1.
We induct in $|\alpha|+k$. The base case is trivial. If we inductively differentiate in $x_i$ where $\beta_i\neq0$ we get
$$(C_1x^{\beta-e_i}u^{-|\beta|/2-|\alpha|/2-k}+ C_2 x^{\beta+e_i} u^{-|\beta|/2-|\alpha|/2-1-k}) p(u,x)$$
which are terms of the correct form
$$x^{\beta’}u^{-|\beta’|/2-(|\alpha|+1)/2-k} p(u,x)$$
with $\beta’=\beta-e_i$ and $\beta’=\beta+e_i$ respectively. The case $\beta_i=0$ is simpler, and differentiating in $u$ instead presents no new difficulties: we get in this case
$$(C_3 x^{\beta}u^{-|\beta|/2-|\alpha|/2-(k+1)} + C_4|x|^2x^{\beta}u^{-|\beta|/2-1-|\alpha|/2-(k+1)}) p(u,x).$$
These are terms of the form
$$x^{\beta''}u^{-|\beta''|/2-|\alpha|/2-(k+1)} p(u,x)$$
with $\beta''=\beta$ or $\beta'' = \beta+2e_i$ with $i=1,\dots,q$, as needed.  So the proof is completed.
